Simplifying $2\cos(t)\cos(2t)-\sin(t)\sin(2t)$ How do I simplify $2\cos(t)\cos(2t)-\sin(t)\sin(2t)$? I know this should be possible, but I don't know how.
I have tried the $\cos(t)\cos(u)-\sin(t)\sin(u)=\cos(t+u)$, but I don't know what to do with the $2$ in front of $\cos(t)$.
 A: $$2\cos(t)\cos(2t)-\sin(t)\sin(2t)=\frac{\cos(t)+3\cos(3t)}{2}$$

Proof:
$$2\cos(t)\cos(2t)-\sin(t)\sin(2t)=\frac{\cos(t)+3\cos(3t)}{2}\Longleftrightarrow$$
$$2\left(2\cos(t)\cos(2t)-\sin(t)\sin(2t)\right)=\cos(t)+3\cos(3t)\Longleftrightarrow$$
$$2\left(\cos(-t)+\cos(3t)-\sin(t)\sin(2t)\right)=\cos(t)+3\cos(3t)\Longleftrightarrow$$
$$2\left(\cos(t)+\cos(3t)-\sin(t)\sin(2t)\right)=\cos(t)+3\cos(3t)\Longleftrightarrow$$
$$2\left(\cos(t)+\cos(3t)+\frac{\cos(3t)-\cos(-t)}{2}\right)=\cos(t)+3\cos(3t)\Longleftrightarrow$$
$$2\left(\cos(t)+\cos(3t)+\frac{\cos(3t)-\cos(t)}{2}\right)=\cos(t)+3\cos(3t)\Longleftrightarrow$$
$$2\left(\cos(t)+\cos(3t)+\frac{\cos(3t)}{2}-\frac{\cos(t)}{2}\right)=\cos(t)+3\cos(3t)\Longleftrightarrow$$
$$2\left(\frac{\cos(t)}{2}+\frac{3}{2}\cos(3t)\right)=\cos(t)+3\cos(3t)\Longleftrightarrow$$
$$\cos(t)+3\cos(3t)=\cos(t)+3\cos(3t)$$
A: It depends how you classify simpler. One possibility is:
$$2\cos t\cos 2t-\sin t\sin 2t=\frac{3}{2}\cos t\cos 2t-\frac{3}{2}\sin t\sin 2t+\frac{1}{2}\cos t\cos 2t+\frac{1}{2}\sin t\sin 2t$$
$$=\frac{1}{2}\big(3(\cos t\cos2t-\sin t\sin 2t)+(\cos t\cos 2t+\sin t\sin 2t)\big)$$
$$=\frac{1}{2}\left(3\cos 3t+\cos t\right)$$
This does make use of the formula you were referencing.
A: You said it yourself, but were confused what to do with the 2:
$$2\cos{(t)}\cos{(2t)}=\cos{(t)}\cos{(2t)}+\cos{(t)}\cos{(2t)}$$
So then your expression becomes
$$\cos{(t)}\cos{(2t)}+\left[\cos{(t)}\cos{(2t)}-\sin{(t)}\sin{(2t)}\right]$$
Now simplify the bracketed expression to $\cos{(3t)}$ by your identity and you have
$$2\cos{(t)}\cos{(2t)}-\sin{(t)}\sin{(2t)}=\cos{(t)}\cos{(2t)}+\cos{(3t)}$$
A: Just have a bit of patience:
\begin{align}
2\cos t\cos2t-\sin t\sin2t
&=2\cos t(2\cos^2t-1)-2\sin^2t\cos t\\
&=2\cos t(2\cos^2t-1)-2\cos t(1-\cos^2t)\\
&=2\cos t(2\cos^2t-1-1+\cos^2t)\\
&=2\cos t(3\cos^2t-2)
\end{align}
If you had a plus, instead of minus, it would be
$$
2\cos t\cos2t+\sin t\sin2t=2\cos^3t
$$
